So I have been trying to learn about entanglement and quantum teleportation and from what I've been able to gather so far, the teleportation part seems to be misleading.

At first I thought that the two particles were of a uniform undetermined state which would collapse probabilistically upon observation.

But the more I read into it, it began to seem like the following unimpressive scenario as it relates to the Schrödinger's cat analogy:

It seems as though a dead cat and a live one are put into two boxes that can only be observed once. When one person gets the live cat, they know the other got the dead cat. They can infer this knowledge instantly, but who cares because it took UPS two days to ship the thing.

If this is the case, why do they call it quantum teleportation and if it isn't the case could someone explain it to me better?

It is not teleportation in the sense that you take a particle and relocate it somewhere else. What really occurs is that you have two particles that are separated, and you can take all of the data from one particle and transfer it to the other particle. So the second particle was just a "host body".
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Chris GerigAug 13 '12 at 2:12

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Why do they call it quantum teleportation? Because it's a cool name.
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ripper234Aug 13 '12 at 13:36

2 Answers
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First: I'd like to discourage you from trying to interpret superpositions in terms of Schrödinger's Cat, when what you're trying to understand involves coherent unitary operations performed on the system. Considering superpositions of states of a massive number of particles makes it difficult to meaningfully consider different bases of measurement, which (among other things) is necessary to describe phenomena such as entanglement. (The fact that it prevents meaningful discussion of coherent evolution is basically the phenomenon of decoherence.) If we were to try and describe teleportation in terms of cats, we'd have to describe it this way:

What is put into the boxes are not live cats and dead cats, nor even two independent indeterminate live/dead cats, but a pair of cats A and B whose states of living are entangled with one another in a single pure state $|\Psi\rangle$ which is a superposition of one live cat and one dead, but where there is no definite fact of whether A is the live one and B the dead one, or vise versa.

The measurement we perform is not just to open a box to see a cat is alive; the pair of entangled cats are nothing but a correlated system which we will use to transmit more information. In this case, we are interested in transmitting yet another cat C, whose state $|\varphi\rangle$ of life/death is also indeterminate, but independent of A and B. We perform a quantum measurement corresponding to disentangling A and C (although they weren't entangled to begin with), and then observe whether they each of A and C is alive or dead (which will cause a collapse of their wavefunctions). Depending on the outcomes of these measurements — A alive and C alive, or A alive and C dead, or etc. — we then perform two operations:

another operation which does, er, something which I can't describe easily in terms of the life and death status of cats. Perhaps it changes the colour of its coat, or something; after all, C's coat may be differently coloured than B's was to begin with.

This is rather unweildly, and as you might imagine, doesn't actually shed any light on the matter. But this isn't surprising — Remember, Schrödinger's Cat was a thought experiment which was designed quite deliberately, by Erwin Schrödinger, to be absurd. If you hope to get any intuition from it, you're almost certainly going to fail, and if you seem to get a simple explanation for what's going on by pretending a quantum system is a cat, you've probably missed some essential details.

Quite seriously, teleportation does involve two elements which are very much quantum: the fact that an entangled resource state is used — two particles A and B, neither of which are the system whose state is being 'teleported', in a joint state $$|\Psi\rangle_{A,B} \;=\; \tfrac{1}{\sqrt 2}\Bigl( |0\rangle_A|1\rangle_B \;-\; |1\rangle_A|0\rangle_B \Bigr) \qquad\quad \Bigr(\text{or a similar state}\Bigr)$$ — and that the state $|\varphi\rangle$ of a third system C can be transmitted to B, by performing a Bell measurement on A and C, which can be simulated by performing the transformation $(H \otimes I \otimes I )(\text{CNOT} \otimes I)$ on the state $|\varphi\rangle_C\otimes|\Psi\rangle_{A,B}$ and measuring $C$ and $A$ in the standard basis, and then performing simple single-qubit operations on B depending on the outcome.

The information about the measurements of A and C have to be transmitted to B somehow; these are sent as classical signals, and travel only at the speed of light. Until they arrive, the system at B is essentially random: whoever had the particle A knows what state it's in, but it may not be the original state of C, and whoever owns B has no useful way of doing anything with it until they know what operations they have to perform to recover the state of C. However, despite the fact that the measurement outcomes on A and C are 'classical', the state which can be recovered with them is fully quantum.

For more complete details about precisely what measurement is performed and how the entangled state is used, you could consult the Wikipedia article on teleportation, this analysis of teleportation elsewhere on this site, or any reasonable text on quantum information (such as Nielsen & Chaung). Ultimately, there's little that you can do but crank through the mathematics, because that is what shows what's going on. In particular, if you want to see how teleportation doesn't make information instantly available at B without communication, you should look at it in terms of what the density operator at B is: up until it obtains the outcome of the measurements, it looks maximally mixed, i.e. totally random.

In any case, that is why it's called quantum teleportation. You could argue about the 'teleportation' part of the name — it's more like a radio transmission the than classic Star Trek style materialization via a beam of energy — but there is no doubts about the quantumness of it.

I'm not satisfied, and have I read the article several times. I may not fully understand bell's inequality (yet), but as far as I can tell, either the state is undermined until it collapses and FTL communication is possible or the state is hidden but deterministic. Your condescending rant has done nothing for me.
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brysgoAug 13 '12 at 17:49

@brysgo: It was meant to be light-hearted, not condescending -- sorry. But as far as FTL communication goes: how is the person on the other side supposed to determine that they go the state that they were supposed to, as opposed to one needing corrections? The potential deviation from the original state of C, due to the possibility that they do not obtain the "preferred" measurement outcome, makes the state B has after measurement indistinguishable from what it was before measurement: that is, totally random noise. Beyond that is a question only of interpretation; but there's no usable signal.
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Niel de BeaudrapAug 13 '12 at 17:54

@brysgo: I've revised the answer a bit to try to clarify things a little. But in terms of the technical aspects of the presentation, I haven't added much. If you want to see how teleportation doesn't make information available at B faster than lightspeed, you should look at the information available at B in terms of density operators.
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Niel de BeaudrapAug 14 '12 at 17:21

Thanks for the clarifications, I am still puzzled, but I will be back after I have done a little more reading on the subject.
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brysgoAug 15 '12 at 16:09

@brysgo: I feel your pain with respect to high-level descriptions of quantum mechanical ideas; they always frustrated me -- ultimately, there's no real recourse but to look at the math. I would ideally like to do exactly that, show you the math; but unfortunately, this involves either quite a high comfort-level tensor products, or $32 \times 32$ matrices. If you had to learn one piece of math to help form intuitions, learn about how products of the form $(U_{C,A} \otimes I_B) \rho_{C,A,B} (U_{C,A}^\dagger \otimes I_B)$ affect measurement statistics on B (where $I$ is the identity).
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Niel de BeaudrapAug 15 '12 at 17:04

To answer to the question should be one of the authors of the original paper. For this reason I post the link to a nice story of Prof. Asher Peres, in which he tells the story of that paper, including how they decide to call it like this.